Let $M$ be a Noetherian $A$-module (where $A$ is a commutative ring).
I want to show that $M$ is faithful implies $A$ is Noetherian.
I am very aware that this question is fully answered in both of these posts:
If $M$ is Noetherian, then $R/\text{Ann}(M)$ is Noetherian, where $M$ is an $R$-module
Let $M$ be a noetherian $R$-module and $I=\mathrm{Ann}_R(M)$. Then $R/I$ is a noetherian ring.
But I would really like to try and get my (naive) method below to work (if possible). So I would be very grateful if anyone could help me make the following idea work...
Suppose $A$ is not Noetherian. Then we can find a chain of ideals
$$I_1\subsetneq I_2 \subsetneq \ldots$$
My idea is to lift this to a strictly increasing chain of submodules of $M.$
Naturally, my first thought is to consider the chain
$$I_1M\subseteq I_2M \subseteq \ldots$$
Can I prove this is strictly increasing?
Well suppose that $I\subsetneq J$ are ideals of $A.$
Take $y \in J\setminus I$ and consider the $A$-linear map $\phi:M\to M$ given by
$$\phi(m)=ym.$$
If $JM=IM,$ then (by the Cayley-Hamilton theorem) we have a relation
$$\phi^n+a_1\phi^{n-1}+\ldots+a_n\mathrm{id}_M=0$$
where the $a_i$ belong to $I.$ Since $M$ is faithful, it follows that
$$y^n+a_1y^{n-1}+\ldots+a_n=0$$
and hence we have shown that $y^n \in I$ which is not quite what we wanted. I then thought perhaps I could take radicals of the original chain of ideals and then apply the above result but of course the process of taking radicals does not necessarily preserve strict inclusion and so I'm a bit stuck now...